Difference between revisions of "2013 USAJMO Problems/Problem 6"
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+ | ==Problem 6== | ||
+ | Find all real numbers <math>x,y,z\geq 1</math> satisfying <cmath>\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.</cmath> | ||
+ | |||
==Solution with Thought Process== | ==Solution with Thought Process== | ||
Without loss of generality, let <math>1 \le x \le y \le z</math>. Then <math>\sqrt{x + xyz} = \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}</math>. | Without loss of generality, let <math>1 \le x \le y \le z</math>. Then <math>\sqrt{x + xyz} = \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}</math>. | ||
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<cmath>b^2 - 4ac = 4(y-1) + 4(z-1) - 8\sqrt{(y-1)(z-1)} - yz(12 + 4yz - 4y - 4z - 8\sqrt{(y-1)(z-1)})</cmath> | <cmath>b^2 - 4ac = 4(y-1) + 4(z-1) - 8\sqrt{(y-1)(z-1)} - yz(12 + 4yz - 4y - 4z - 8\sqrt{(y-1)(z-1)})</cmath> | ||
− | < | + | Let us try a few examples. If y = z, then the discriminant D = <math>8(y-1) - 8(y-1) - yz(12 + 4y^2 - 8y - 8(y-1)) = -yz(4y^2 - 16y + 20) = -4yz(y^2 - 4y + 5) < 0</math>. |
− | Success! The discriminant is negative. Thus, we can replace our claim with a strict one, and there are | + | |
+ | We are almost done, but we need to find the correct argument. (How frustrating!) | ||
+ | Success! The discriminant is negative. Thus, we can replace our claim with a strict one, and there are no real solutions to the original equation in the hypothesis. | ||
--Thinking Process by suli | --Thinking Process by suli |
Latest revision as of 20:02, 30 April 2014
Problem 6
Find all real numbers satisfying
Solution with Thought Process
Without loss of generality, let . Then
.
Suppose x = y = z. Then , so
. It is easily verified that
has no solution in positive numbers greater than 1. Thus,
for x = y = z. We suspect if the inequality always holds.
Let x = 1. Then we have , which simplifies to
and hence
Let us try a few examples: if y = z = 2, we have
; if y = z, we have
, which reduces to
. The discriminant (16 - 20) is negative, so in fact the inequality is strict. Now notice that yz - y - z + 3 = (y-1)(z-1) + 2. Now we see we can let
! Thus,
and the claim holds for x = 1.
If x > 1, we see the will provide a huge obstacle when squaring. But, using the identity
:
which leads to
Again, we experiment. If x = 2, y = 3, and z = 3, then
.
Now, we see the finish: setting gives
. We can solve a quadratic in u! Because this problem is a #6, the crown jewel of USAJMO problems, we do not hesitate in computing the messy computations:
Because the coefficient of is positive, all we need to do is to verify that the discriminant is nonpositive:
Let us try a few examples. If y = z, then the discriminant D = .
We are almost done, but we need to find the correct argument. (How frustrating!) Success! The discriminant is negative. Thus, we can replace our claim with a strict one, and there are no real solutions to the original equation in the hypothesis.
--Thinking Process by suli